Question

Point charges q1=+2.00μc and q2=−2.00μc are placed at adjacent corners of a square for which the length of each side is 1.50 cm . point a is at the center of the square, and point b is at the empty corner closest to q2. take the electric potential to be zero at a distance far from both charges. part a what is the electric potential at point a due to q1 and q2? express your answer with the appropriate units.

Answer 1

Explanation :

It is given that,

$q_1=+2\mu C=+2\times 10^{-6}\ C$

$q_2=-2\mu C=-2\times 10^{-6}\ C$

Side of square, $x=1.5\ cm=0.015\ m$

It is given that, the potential at point c is equal to zero.

So, potential at point a due to $q_1$ and  $q_2$ is given as:

$V_a=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}$

where, $r_1=r_2=\dfrac{1}{\sqrt{2} }(0.015\ m)$

So,  $V=\dfrac{kq_1}{r_1}+(-\dfrac{kq_2}{r_2})$  (Since, $q_2$ is negative)

Hence, the potential at a due to $q_1$ and  $q_2$ is equal to 0 V.